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December 30th, 2018, 11:48 PM   #11
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 Originally Posted by Integrator Hello, The problem I posted is from another forum...It is known that if $\displaystyle \overrightarrow{z}=\overrightarrow{x}+ \overrightarrow{y}$ , then $\displaystyle \overrightarrow{z} ^2=\overrightarrow{x}^2+\overrightarrow{y}^2+2 \overrightarrow{x}\overrightarrow{y}$ and so what expressions do they have $\displaystyle \overrightarrow{z} ^3$ , $\displaystyle \overrightarrow{z}^4$ , $\displaystyle \overrightarrow{z}^5$ ,......., $\displaystyle \overrightarrow{z}^{n-1}$ , $\displaystyle \overrightarrow{z}^n$? Thank you very much! All the best! Integrator
Then ask the other forum whether they mean the geometric algebra product, or just the dot product. If they mean just the dot product, then tell the other forum that the question as stated does not make any sense (in standard linear algebra). December 31st, 2018, 12:22 PM   #12
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 Originally Posted by Integrator Hello, The problem I posted is from another forum...It is known that if $\displaystyle \overrightarrow{z}=\overrightarrow{x}+ \overrightarrow{y}$ , then $\displaystyle \overrightarrow{z} ^2=\overrightarrow{x}^2+\overrightarrow{y}^2+2 \overrightarrow{x}\overrightarrow{y}$ and so what expressions do they have $\displaystyle \overrightarrow{z} ^3$ , $\displaystyle \overrightarrow{z}^4$ , $\displaystyle \overrightarrow{z}^5$ ,......., $\displaystyle \overrightarrow{z}^{n-1}$ , $\displaystyle \overrightarrow{z}^n$? Thank you very much! All the best! Integrator
Again, $\displaystyle (\vec{z})^2 = \vec{z} \cdot \vec{z}$ is a scalar. I suppose you can calculate $\displaystyle (\vec{z} )^3 = ( \vec{z} \cdot \vec{z} ) (\vec{z}) = ( \vec{z} )^2 ( \vec{x} + \vec{y} )$ but I don't see the point to it. I would think that we'd like to have $\displaystyle (\vec{z})^n$ to be a vector for all n. You can't do that with the dot product or with the cross product.

-Dan

Last edited by topsquark; December 31st, 2018 at 12:26 PM. December 31st, 2018, 12:36 PM   #13
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 Originally Posted by topsquark Again, $\displaystyle (\vec{z})^2 = \vec{z} \cdot \vec{z}$ is a scalar. I suppose you can calculate $\displaystyle (\vec{z} )^3 = ( \vec{z} \cdot \vec{z} ) (\vec{z}) = ( \vec{z} )^2 ( \vec{x} + \vec{y} )$ but I don't see the point to it. I would think that we'd like to have $\displaystyle (\vec{z})^n$ to be a vector for all n. You can't do that with the dot product or with the cross product. Do you have any more information about the problem? -Dan
Can't you do it with the cross product? (cross vector being a pseudovector aside) What am I missing? December 31st, 2018, 01:29 PM   #14
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 Originally Posted by Micrm@ss Can't you do it with the cross product? (cross vector being a pseudovector aside) What am I missing?
As Romsek pointed out in post 6: $\displaystyle \vec{z} \times \vec{z} = 0$, so $\displaystyle \vec{z} ^n = \vec{z} \times \vec{z} \times \text{ ...}$ where n is greater than 1 is simply the 0 vector. Nothing interesting about doing it this way.

-Dan December 31st, 2018, 02:44 PM   #15
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 Originally Posted by topsquark As Romsek pointed out in post 6: $\displaystyle \vec{z} \times \vec{z} = 0$, so $\displaystyle \vec{z} ^n = \vec{z} \times \vec{z} \times \text{ ...}$ where n is greater than 1 is simply the 0 vector. Nothing interesting about doing it this way. -Dan
Sure, but not impossible to define it. Just not interesting. The two are different Like my girlfriend, she is impossible, but interesting... December 31st, 2018, 03:54 PM   #16
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 Originally Posted by Micrm@ss Like my girlfriend, she is impossible, but interesting...
Oooooh are you going to get in trouble!

-Dan January 1st, 2019, 01:46 AM   #17
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 Originally Posted by topsquark Oooooh are you going to get in trouble! -Dan
Not impossible January 1st, 2019, 01:01 PM   #18
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 Originally Posted by Integrator Hello, The problem I posted is from another forum...It is known that if $\displaystyle \overrightarrow{z}=\overrightarrow{x}+ \overrightarrow{y}$ , then $\displaystyle \overrightarrow{z} ^2=\overrightarrow{x}^2+\overrightarrow{y}^2+2 \overrightarrow{x}\overrightarrow{y}$ and so what expressions do they have $\displaystyle \overrightarrow{z} ^3$ , $\displaystyle \overrightarrow{z}^4$ , $\displaystyle \overrightarrow{z}^5$ ,......., $\displaystyle \overrightarrow{z}^{n-1}$ , $\displaystyle \overrightarrow{z}^n$? Thank you very much! All the best! Integrator
For an exponent of 2, the product is a dot product. It is not defined for higher exponents. January 1st, 2019, 10:09 PM   #19
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 Originally Posted by mathman For an exponent of 2, the product is a dot product. It is not defined for higher exponents.
Hello,

Why can not be defined for $\displaystyle n>2$?
However, some claim that according to Newton's binomial $\displaystyle \overrightarrow{z}^n=\overrightarrow{x}^n+C_n^1 \cdot \overrightarrow{x}^{n-1}\cdot \overrightarrow{y}+\cdots +C_n^{n-1}\cdot \overrightarrow{x}\cdot \overrightarrow{y}^{n-1}+\overrightarrow{y}^n$.

All the best,

Integrator January 1st, 2019, 10:43 PM   #20
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 Originally Posted by Integrator However, some claim that according to Newton's binomial $\displaystyle \overrightarrow{z}^n=\overrightarrow{x}^n+C_n^1 \cdot \overrightarrow{x}^{n-1}\cdot \overrightarrow{y}+\cdots +C_n^{n-1}\cdot \overrightarrow{x}\cdot \overrightarrow{y}^{n-1}+\overrightarrow{y}^n$.
They are wrong. What you wrote makes no sense at all in standard linear algebra. Tags calculation, vectorial Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post EBTERTTBT Physics 5 August 14th, 2018 05:29 AM markovski Math 0 May 22nd, 2017 09:17 AM juanpe966 Calculus 4 February 27th, 2016 02:24 PM joskevermeulen Calculus 1 December 29th, 2015 05:02 AM FunWarrior Linear Algebra 0 February 11th, 2014 01:22 AM

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